Having function v(x,a) - velocity for simple harmonic movement - as described below:

enter image description here

If I do the summation like this:

enter image description here

The result is 31.97. I would expect it as 32 instead. Why I still get the approximated result? If I keep increasing the a parameter the summation becomes more and more away from 32.


I realized I didn't give too much context of what I am trying to achieve, so here it is:

I am trying to simulate a simple harmonic movement (a platform going up and down, repeatedly), so I thought would be a good idea to use sine/cosine functions to go about this.

So my constraint for a graph of Position x Frames (time) would look like this:

enter image description here

enter image description here

As you can see, the maximum amplitude point of this wave is between 32 and -32.

But I simply cannot assign the position directly (limitations of the program I am using), instead, I have to set the velocity. The velocity is always 60x the position. That means if I assign 1 to the velocity during 1 frame, the displacement will be 1/60 = 0.01666666666666667. 60 is the number of frames per second.

So the first function v(x, a) was my first attempt to make sure, after N cycles the maximum displacement would be always 32. So I don't know what function I should use to have a result displacement (summation) of 32.

I hope I made my question a bit more clear, thanks for relentless help.

  • 7
    $\begingroup$ Why would you expect it to be $32$? $\endgroup$ – Matt Samuel Nov 8 '19 at 12:35
  • 3
    $\begingroup$ On the other hand, we can be quite sure that $\sum_{n=1}^{30}\frac{\nu(n,2)}{60}=32\frac{\pi}{60}\sum_{k=1}^{30}\sin\frac{n\pi }{30}$ is a multiple of $\pi$ by the algebraic number $\frac{32}{60}\sum_{n=1}^{30}\sin\frac{n\pi}{30}$, and therefore it is not $32$. $\endgroup$ – Gae. S. Nov 8 '19 at 12:38
  • $\begingroup$ Have a look at my edit. $\endgroup$ – Claude Leibovici Nov 10 '19 at 4:32

You are facing the sum of sines where the angles are in arithmetic progression. If you apply the formula, you will notice that $$\frac{32\pi}{60}\sum_{k=1}^{30}\sin\left(\frac{n\pi }{30}\right)=\frac{8\pi}{15} \cot \left(\frac{\pi }{60}\right)$$ We do not know the values of the trigonometric function for such an angle.

Since te argument is small, let us use the usual Taylor series $$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2 x^5}{945}+O\left(x^7\right)$$ Replacing $x$ by $\frac{\pi }{60}$ will lead to $$32-\frac{2 \pi ^2}{675}-\frac{\pi ^4}{18225000}-\frac{\pi ^6}{688905000000}$$


Considering that we look for

$$S(a)=\frac{4}{15} \pi a\sum_{x=1}^{30}\sin \left(\frac{\pi a x}{60}\right)=\frac{2}{15} \pi a \left(\cot \left(\frac{\pi a}{120}\right)-\cos \left(\frac{61 \pi a}{120}\right) \csc \left(\frac{\pi a}{120}\right)\right)$$ and that we want the result to be as close as possible to $32$,we can minimize $$\Phi(a)=|S(a)-32|$$ which leads to $$a_{opt}=1.96703696\qquad \text{and} \qquad S(a_{opt})=31.99292724$$

| cite | improve this answer | |
  • $\begingroup$ Great, if I need to find a value for "a" that result in a sum result of 32, how would I go about it? $\endgroup$ – raphaklaus Nov 8 '19 at 14:56
  • $\begingroup$ @raphaklaus. Could you provide the exact summation for the general case ? Is it always from $1$ to $30$ ? Is the result always divided by $60$ ? $\endgroup$ – Claude Leibovici Nov 8 '19 at 15:11
  • $\begingroup$ Sorry, I didn't give the context. I am simulating a simple harmonic movement with this function. The first function I tried to calculate the velocity in a way that the maximum position will be always 32, but found out that it was not right. I divide by 60 in the summation because it is the number of frames per second. Every result of the velocity divided by 60 is the amount of displacement in the Y axis is going to happen per frame. $\endgroup$ – raphaklaus Nov 9 '19 at 9:19
  • $\begingroup$ @raphaklaus. You did not answer my question. $\endgroup$ – Claude Leibovici Nov 9 '19 at 9:22
  • $\begingroup$ 1 to 30 because 30 represents the Period (T). Always divided by 60 because I have to know the total displacement per frame. $\endgroup$ – raphaklaus Nov 9 '19 at 11:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.